Rank and submatrices theorem

  • Thread starter Grothard
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  • #1
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Main Question or Discussion Point

Let A be a nonzero matrix of size n. Let a k*k submatrix of A be defined as a matrix obtained by deleting any n-k rows and n-k columns of A. Let m denote the largest integer such that some m*m submatrix has a nonzero determinant. Then rank(A) = k.

Conversely suppose that rank(A) = m. There exists a m*m submatrix has a nonzero determinant.



I'm currently trying to prove this theorem. Not quite sure if I should proceed by examining the solution space of A or rather just do something clever with the determinants. I feel like there's a property of determinants that I'm missing that'd make this much easier.
 

Answers and Replies

  • #2
1,761
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If the determinant is non-zero, that implies that the columns are linearly independent. Remember, the determinant measures the (hyper-)volume of the parallelepiped spanned by the columns. Intuitively, a non-zero volume implies linear independence. I would use that.

So, you have a square submatrix whose columns are linearly independent. What does that tell you about the corresponding columns in the bigger matrix A?
 
  • #3
Deveno
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some things to think about:

if you use row-reduction to find rank(A), how does each row-operation affect the determinant?

if you think of the determinant as a function of n n-vectors, instead of an nxn matrix, is it linear in each variable? how does this tie in to dim(row(A)) = dim(col(A))?

does re-arranging rows or columns of a matrix change its determinant?
 

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